Find the height of the square pyramid
slachterman
triangular face makes with the base are in a ratio of 1:2:2:4. Find
the height h.
LudovicoVan
The height h is given.
-LV
slachterman
Hilarious. h denotes the length of the line. What is the length of the
line?
LudovicoVan
> On Mar 12, 2:52 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > On Mar 12, 7:37 pm, slachterman <slachter...@gmail.com> wrote:
>
> > > Given a square pyramid of height h, the (dihedral) angles that each
> > > triangular face makes with the base are in a ratio of 1:2:2:4. Find
> > > the height h.
>
> > The height h is given.
>
> line?
As far as I can tell, the problem is ill stated: you should give
something else in function of which h should be calculated (e.g. the
length of the side of the base, otherwise a parameter denoting it, or
similar). The ratio of the angles alone is not enough.
-LV
LudovicoVan
Besides, I was actually thinking of the ratios 1:2:4:2. I am not even
sure the ratios you give can actually work out to a pyramid at all.
-LV
Ray Vickson
1) Hilarious. Who says it is even possible to have a square pyramid
with those angle ratios?
2) Hilarious. If the diameter of the base is 1 cm the height h will be
different than if the diameter of the base is 1,000,000,000 cm.
R.G. Vickson
rasterspace
a)
with a given area of the tetragonal base,
the four facets may not intersect at a single vertex. so, b)
David Hartley
<8cc885ce-395d-49ea...@f36g2000pri.googlegroups.com>, Ray
Vickson <RGVi...@shaw.ca> writes
1) It may not be possible in general, but it is with these particular
ratios.
2) Just assume the OP meant find the ratio between the height and the
side of the base.
--
David Hartley
slachterman
> In message
> <8cc885ce-395d-49ea-8e2d-c9070a197...@f36g2000pri.googlegroups.com>, Ray
> Vickson <RGVick...@shaw.ca> writes
Assume the side of the base has length one.
LudovicoVan
> In message
> <8cc885ce-395d-49ea-8e2d-c9070a197...@f36g2000pri.googlegroups.com>, Ray
> >On Mar 12, 1:41 pm, slachterman <slachter...@gmail.com> wrote:
> >> On Mar 12, 2:52 pm, LudovicoVan <ju...@diegidio.name> wrote:
> >> > On Mar 12, 7:37 pm, slachterman <slachter...@gmail.com> wrote:
>
> >> > > Given a square pyramid of height h, the (dihedral) angles that each
> >> > > triangular face makes with the base are in a ratio of 1:2:2:4. Find
> >> > > the height h.
>
> >> > The height h is given.
>
> >> line?
>
> >1) Hilarious. Who says it is even possible to have a square pyramid
> >with those angle ratios?
> >2) Hilarious. If the diameter of the base is 1 cm the height h will be
> >different than if the diameter of the base is 1,000,000,000 cm.
>
> 1) It may not be possible in general, but it is with these particular
> ratios.
> 2) Just assume the OP meant find the ratio between the height and the
> side of the base.
Given your comment, I have tried: there is no solution, unless you
allow h=0 or I have made a mistake.
-LV
David Hartley
<920b0652-f8ae-4539...@u14g2000vbg.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> writes
Allow the largest angle to be greater than 90 degrees; one side leaning
away from the base.
--
David Hartley
rasterspace
could make an easy solution perhaps both
for 1:2:2:4 AND 1:2:4:2, taking the edges
cyclically, althoug htey may work, without that.
LudovicoVan
I have not imposed any constraint on the angles.
-LV
David Hartley
<a7da9ecb-9fb4-492b...@q12g2000prb.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> writes
>> Allow the largest angle to be greater than 90 degrees; one side leaning
>> away from the base.
>
>I have not imposed any constraint on the angles.
Perhaps you're taking the angles in a different order. I was assuming
the two equal angles were opposite each other, so basically you want a
solution of
cot(A) + cot(4A) = 2cot(2A)
With the other order you'd need
cot(A) + cot(2A) = cot(2A) + cot(4A)
whose only solution is, as you said, A = 0.
--
David Hartley
LudovicoVan
Right: but for the fact that the one who's taken the angles in
different order is you! Plus you've just snipped the whole problem
statement... Bad bad bad, you get an H on the whole line.
-LV
LudovicoVan
Anyway, nice to see your approach: I had used tangents instead, and
the whole thing was much more convoluted. Here it is, in case someone
has any specific comments or corrections:
>>>>>>> On Mar 12, 7:37 pm, slachterman <slachter...@gmail.com> wrote:
>>>>>>> > Given a square pyramid of height h, the (dihedral) angles that each
>>>>>>> > triangular face makes with the base are in a ratio of 1:2:2:4. Find
>>>>>>> > the height h [in terms of the side of the base].
(In my tentative mathematical English:)
Let H be the vertex of the pyramid, and O its perpendicular to the
base. Let A, D, C, D be the perpendiculars of H to the sides of the
base along the triangular faces, and a, 2a, 2a, 4a the values of
angles HAO, HBO, HCO, HDO respectively.
Let l be the length of the side of the base. Denote by x the length
of AO, and by y the length of CO. Then:
length(AO) = x
length(BO) = y
length(CO) = l-x
length(DO) = l-y
Consider the right triangles HOA, HOC, HOB, HOD. Then:
h = x tan(a) (1)
h = y tan(2a) (2)
h = (l-x) tan(2a) (3)
h = (l-y) tan(4a) (4)
Work out x, y from (1), (2):
x = h/tan(a) (1')
y = h/tan(2a) (2')
Substitute (1'), (2') in (3), (4):
h = (l - h/tan(a)) tan(2a) (3')
h = (l - h/tan(2a)) tan(4a) (4')
Work out h from (3'), (4'):
h = l (tan(a) tan(2a)) / (tan(a) + tan(2a)) (3'')
h = l (tan(2a) tan(4a)) / (tan(2a) + tan(4a)) (4'')
Divide (3'') by (4'') to work out a:
(tan(a) tan(2a)) / (tan(a) + tan(2a))
1 = --------------------------------------- (5)
(tan(2a) tan(4a)) / (tan(2a) + tan(4a))
Simplify (5) -- here omitting intermediate passages:
tan^2(a) (tan^2(a) - 3) (tan^2(a) + 1) = 0 (5')
We get three cases:
tan^2(a) = 0 (5'.1)
tan^2(a) = 3 (5'.2)
tan^2(a) = -1 (5'.3)
That is -- omitting negative solutions due to symmetry:
tan(a) = 0 (5'.1')
tan(a) = sqrt(3) (5'.2')
tan(a) = sqrt(-1) (5'.3') IMPOSSIBLE
From (5'.1'), (5'.2'):
a = 0 (5'.1'')
a = pi/3 (5'.2'')
Case (5'.1'') corresponds to the degenerate pyramid with a=0 and h=0.
Case (5'.2'') turns out to be IMPOSSIBLE as well as soon as we try and
substitute in either of (3'') and (4''), because we get zero at the
denominators.
Bottom line, there is no solution other than the degenerate one given
by (5'.1'').
-LV
LudovicoVan
> Denote by x the length
> of AO, and by y the length of CO.
Typo: should read "y the length of BO".
> Consider the right triangles HOA, HOC, HOB, HOD.
Typo: should read "the right triangles HOA, HOB, HOC, HOD".
-LV
Philippe 92
> On Mar 15, 11:44 am, LudovicoVan <ju...@diegidio.name> wrote:
>> On Mar 15, 11:28 am, David Hartley <m...@privacy.net> wrote:
>>> cot(A) + cot(4A) = 2cot(2A)
>>>
>>> With the other order you'd need
>>> cot(A) + cot(2A) = cot(2A) + cot(4A)
>>> whose only solution is, as you said, A = 0.
>>
>
> the whole thing was much more convoluted. Here it is, in case someone
> has any specific comments or corrections:
>
>>>>>>>> On Mar 12, 7:37 pm, slachterman <slachter...@gmail.com> wrote:
>>>>>>>>> Given a square pyramid of height h, the (dihedral) angles that each
>>>>>>>>> triangular face makes with the base are in a ratio of 1:2:2:4. Find
>>>>>>>>> the height h [in terms of the side of the base].
>
> h = (l - h/tan(a)) tan(2a) (3')
> h = (l - h/tan(2a)) tan(4a) (4')
>
These equations may be written
h/tan(2a) + h/tan(a) = 1 (3bis)
and
h/tan(4a) + h/tan(2a) = 1 (4bis)
hence :
h/tan(2a) + h/tan(a) = h/tan(4a) + h/tan(2a)
that is
cot(2a) + cot(a) = cot(4a) + cot(2a),
exactly the equation given by David in this case 1:2:2:4
that is cot(a) = cot(4a), the only solution with 0 < 4a < 180 deg
being a = 0 (hum...)
The other equation (in the case 1:2:4:2) is similarily obtained.
But the case 1:2:4:2 gives a nice solution...
So may be there is a mistake in the original post by slachterman !
(the 1:4 being at opposite base sides, instead of adjacent sides, so
that the problem does have some interest)
Regards.
--
Philippe C., mail : chephip, with domain free.fr
site : http://mathafou.free.fr/ (mathematical recreations)